Page 176 - Vogel's TEXTBOOK OF QUANTITATIVE CHEMICAL ANALYSIS
P. 176

4   ERRORS AND STATlSTlCS

          Total sum of  squares = (0.0269 + 0.0157 + 0.0539) - C.F.
                             = 0.0965 - 0.0234 = 0.073 1
       (d)  The between-treatment (analyst) surn of  squares. The sum of  the squares of
          each individual column is divided by the number of results in each column,
          and then the correction factor is subtracted.
          Between sum of  squares = 6(0.292 + 0.21  + 0.452) - 0.0234
                                = 0.0593
       (e)  The within-sample surn of  squares. The between sum of squares is subtracted
          from the total sum of  squares.
          0.0731 - 0.0593 = 0.0138
       (f)  The degrees of  freedom  (v). These are obtained as follows:
          The total number of  degrees of freedom v = N - 1 = 11
          The between-treatment  degrees of freedom v = C - 1 = 2
          The within-sample  degrees of freedom v = (N - 1 ) - ( C - 1 ) = 9
          where C is the number of columns (in this example, the number of analysts).
       (g) A table of  Analysis of  Variance (ANOVA table) may now be set up.

       Source of variation   Sum of squares   d.f.   Mean square
       'Between analysts'   0.0593   2    0.0593/2 = 0.0297
       'Within titrations'   0.0138   9   0.0138/9 = 0.00153

       Total           0.073 1       11

       (h) The F-test is used to compare the two mean squares:




          From the F-tables (Appendix 13), the value of F at the  1 per cent level for
          the given degrees of freedom is 8.02. The calculated result (19.41) is higher
          than 8.02; hence there is a significant difference in the results obtained by
          the three analysts. Having ascertained in this example there is a significant
          difference between the three analysts, the next stage would be to determine
          whether  the  mean  result  is  different  from  the  others,  or whether  al1 the
          means are significantly different from each other.
         The  procedure  adopted  to  answer  these  questions for  the  example  given
       above is as follows:
       (a) Calculate  the titration means for each analyst. The mean titration  values
          are X (A) = 22.57 mL; X (B) = 22.45 mL; and X (C) = 22.61 mL.
       (b) Calculate the quantity defined as the 'least  significant difference',  which is
                      -
          given  by  s J2/n  to.o,  where  s  is  the  square  root  of  the  Residual  Mean
           Square, Le.  the 'within-titration'  Mean Square. Hence s = dm; is
                                                                       n
           the number of results in each column (in this example, 4); t is the 5 percent
           value from the t-tables (Appendix 12), with the same number of degrees of
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